Show $Q$ with the given addition and scalar multiplication is vector space where $Q = \{v+ H\}$

Let $$H$$ be a subspace of $$V$$.

For $$c\in V,$$ define $$E(c) = \{c + h\,|\,h\in H\}$$

Let $$Q = \{ E(v)\,|\,v\in V\}$$.

Define addition in $$Q$$ by: for $$v, w\in V$$, $$E(v) \bigoplus E(w) = E(v+w)$$

Define scalar multiplication in $$Q$$ by: for $$v\in V$$ and $$\propto \in \mathbb{R}, \propto E(v) = E(\propto v)$$.

Show that $$Q$$ together with this addition and scalar multiplication is a vector space.

My Attempt:

To check that $$Q$$ is a vector space, one must check each of the 10 axioms of a vector space to see if they hold.

$$A_1:$$ Let $$a, b\in Q$$. Then \begin{align*} a + b = E(a) + E(b) \in Q \end{align*} Therefore $$Q$$ is closed under addition ($$A_1$$ holds).

I haven't completed the whole solution yet, but you could imagine how long it would take. Is there a shorter and better way of proving?

• I don't think there is a shorter way. On the other hand, I think all of the axioms are similarly short. It's not really that much work. Jun 2, 2021 at 11:16

You can see $$Q$$ as the subspace of the vector space $$V$$: Using your definition, we have $$Q=\{v+H:v\in V\}$$. This is nothing else than the quotient space (that's why the letter 'Q' is used :D) $$Q=V/H$$ with equivalence classes $$[v]=v+H=\{v+h:h\in H\}$$.

Let's take the direct sum $$V=H\oplus H^\perp$$ , then the quotient space is naturally isomorphic to the orthogonal complement of $$H$$: $$Q=V/H\simeq H^\perp$$.

This shows already, that $$Q$$ is isomorphic to a subspace of $$V$$ and thus a subspace itself.

See https://en.wikipedia.org/wiki/Quotient_space_(linear_algebra) for more details about Quotients of vector spaces.

Edit: If you really want to calculate the vector space axioms you can use the fact that the $$+$$ and $$\cdot$$ operation are inherited from $$V$$ and thus you only need to verify that $$Q$$ is a subspace. For this you only need to verify two axioms

1. $$Q\neq \emptyset$$
2. $$\forall [v],[w]\in Q,\lambda\in\mathbb{R}$$ we have $$[v]+\lambda [w]\in Q$$

and that should be easy.

• Thank you for the informative answer. Just want to ask how do you exactly verify the second axiom you presented?
– muw
Jun 2, 2021 at 11:34
• Instead of $E(v)$ i prefer writing $[v]$. We then have $[v]+\lambda [w] = [v] + [\lambda w] = [v+\lambda w]$ using the addition and multiplication you had defined above. So $[v+\lambda w]$ is also an equivalence class and thus in $Q$ as well Jun 2, 2021 at 11:35
• I see. The notation confused me. Thanks!
– muw
Jun 2, 2021 at 11:36